Question

If n resistors of unequal resistances are connected in parallel,and the percenrage error in all resistors are k then,total error in parallel combination is?

(a) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)

(b) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=1-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)

(c) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=1+\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)

(d) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)

The question was asked by my school teacher while I was bunking the class.

Origin of the question is Errors and Approximations in section Partial Differentiation of Engineering Mathematics

Answer 1

The correct answer is (b) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=1-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)

The explanation: Given  ^1⁄r = ^1⁄a + ^1⁄b + ^1⁄c +⋯.. + ^1⁄n

Differentiating all,

– ^1⁄r^2  dr = – ^1⁄a^2  da – ^1⁄b^2  db – ….- ^1⁄n^2  dn

Now,

^1⁄r^2  dr = + ^1⁄a^2  da + ^1⁄b^2  db + ….+ ^1⁄n^2  dn

Multiplying by 100 and putting da = db = ……. = dn =k.